Formula Mean Values in Matlab code.

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aggelos
aggelos 2012 年 9 月 29 日
Hello, can anyone help to write in Matlab code the formula below. The third term with the minus sine is a double sum.
(sum from i=1 to N of Xt-i*168)/N + (sum from j=1 to 7 of Xt-j*24)/7 -(sum of i=1 to n sum of j=1 to 7 of Xt-i*168-j*24)/(7*N).
thank you for your time
  4 件のコメント
Azzi Abdelmalek
Azzi Abdelmalek 2012 年 9 月 29 日
what are your index i or t?
aggelos
aggelos 2012 年 9 月 29 日
I have the code for a similar formula but i can;t understand it . Here it is. The formula is (sum from i=1 to N of Xt-i*7)/N + (sum from j=1 to 7 of Xt-j7)/7 -(sum of i=1 to n sum of j=1 to 7 of Xt-i*7-j)/(7*N). and the code is
mean(L(start+i-7:-7:start+i-N*7)) + ... mean(L(start+i-1:-1:start+i-7)) - mean(L(start+i-7-1:-1:start+i-(N+1)*7));

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回答 (1 件)

Image Analyst
Image Analyst 2012 年 9 月 29 日
I don't think you've explained this correctly. What's the difference between t and i? Anyway, here's some code to get you started:
N = 100; % Or whatever. The number of elements.
T = 1 : N;
Xt = 2 * T; % Or whatever function of T that X is.
i = 1 : N;
theFirstSum = sum(Xt - (i * 168) / N)
  5 件のコメント
aggelos
aggelos 2012 年 9 月 30 日
Dear Image Analyst, To make the things clear. Forget The 't' index.Let's say that X is my time series (L in the code above). My problem is how to define the indexes in each sum. For example, in the first sum the indexes are from i=1 to N , but i need to take values of the form : for i=1 I need to take the price which is 168 hours before. for i =2 the price which is 2*168=336 hours before ... for i=N the price which is N*168 hours before. Sorry to bother you and thank you very much for your time and interest.
Image Analyst
Image Analyst 2012 年 9 月 30 日
To sum the prior 168 items, in general
theSum = zeros(1, lastIndex);
for k = 168 : lastIndex
theSum(k) = = sum(Xt((k - 167) : k));
end

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